project_euler.problem_027.sol1
==============================

.. py:module:: project_euler.problem_027.sol1

.. autoapi-nested-parse::

   Project Euler Problem 27
   https://projecteuler.net/problem=27

   Problem Statement:

   Euler discovered the remarkable quadratic formula:
   n2 + n + 41
   It turns out that the formula will produce 40 primes for the consecutive values
   n = 0 to 39. However, when n = 40, 402 + 40 + 41 = 40(40 + 1) + 41 is divisible
   by 41, and certainly when n = 41, 412 + 41 + 41 is clearly divisible by 41.
   The incredible formula  n2 - 79n + 1601 was discovered, which produces 80 primes
   for the consecutive values n = 0 to 79. The product of the coefficients, -79 and
   1601, is -126479.
   Considering quadratics of the form:
   n² + an + b, where |a| < 1000 and |b| < 1000
   where |n| is the modulus/absolute value of ne.g. |11| = 11 and |-4| = 4
   Find the product of the coefficients, a and b, for the quadratic expression that
   produces the maximum number of primes for consecutive values of n, starting with
   n = 0.



Functions
---------

.. autoapisummary::

   project_euler.problem_027.sol1.is_prime
   project_euler.problem_027.sol1.solution


Module Contents
---------------

.. py:function:: is_prime(number: int) -> bool

   Checks to see if a number is a prime in O(sqrt(n)).
   A number is prime if it has exactly two factors: 1 and itself.
   Returns boolean representing primality of given number num (i.e., if the
   result is true, then the number is indeed prime else it is not).

   >>> is_prime(2)
   True
   >>> is_prime(3)
   True
   >>> is_prime(27)
   False
   >>> is_prime(2999)
   True
   >>> is_prime(0)
   False
   >>> is_prime(1)
   False
   >>> is_prime(-10)
   False


.. py:function:: solution(a_limit: int = 1000, b_limit: int = 1000) -> int

   >>> solution(1000, 1000)
   -59231
   >>> solution(200, 1000)
   -59231
   >>> solution(200, 200)
   -4925
   >>> solution(-1000, 1000)
   0
   >>> solution(-1000, -1000)
   0